Using a$2-$year moving$$ average, the forecast for year $6=$

$$

$3,825$ miles $($round your response to two decimal$$ places$).$

Part $5$

$$b$)$ Find the MAD based on the$2-$year moving average forecast in part$($a$).$

The value of MAD is computed by taking the sum of the absolute values of the individual forecast errors$($deviations$)$ and dividing by the number of periods of data$($n$)$:

MAD$=$ StartFraction Summation from nothing to nothing $|$Actual minus Forecast$|$ Over n EndFraction

Part $6$

The absolute error for year $4$ is

$$

$125\mathrm{.}00$ miles$($round your response to two decimal$$ places$).$

Part $7$

If a$2-$year moving average is used to make the$$ forecast, the MAD based on this$=$

$$

$150\mathrm{.}0$ miles $($round your response to two decimal$$ places$).($Hint: You will have only $3$ years of matched$$ data.$)$

Part $8$

$$c$)$ Use a weighted$2-$year moving average with weights of $0\mathrm{.}45$ and $0\mathrm{.}55$ to forecast next$$ year's mileage.$($The weight of $0\mathrm{.}55$ is for the most recent$$ year.$)$

A weighted moving average may be expressed mathematically$$ as:

Weighted moving average$=$ StartFraction Upper Sigma left parenthesis Weight for period n right parenthesis left parenthesis Demand in period n right parenthesis Over Upper Sigma Weights EndFraction

Part $9$

For the next$$ year's mileage, the weighted$2-$year moving average is

$$

$.55$times$3$ comma $850$plus

$$

$.45$times$3$ comma $800($round your responses to two decimal$$ places$).$

Part $10$

$$Therefore$,$ the forecast for year $6$ is

$$

$3,823$ miles$($round your response to two decimal$$ places$).$

Part $11$

What MAD results from using this approach to$$ forecasting?

The MAD for the forecast developed using a weighted$2-$year moving average with weights of $0\mathrm{.}45$ and $0\mathrm{.}55=$

$$

enter your response here miles $($round your response to two decimal$$ places$).($Hint: You will have only $3$ years of matched$$ data.$)$